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Home » Triangles » Converse Triangle Midsegment Theorem

Converse Triangle Midsegment Theorem

Last updated: Oct 1, 2019 by Ido Sarig · This website generates income via ads and uses cookies · Terms of use · Privacy policy

In today's geometry lesson, we will prove the Converse Triangle Midsegment Theorem.

The Triangle Midsegment Theorem states that the midsegment of a triangle is parallel to the third side, and its length is equal to half the length of the third side.

We will now prove the converse of this theorem - that if a line connecting two sides of a triangle is parallel to the third side and equal to half that side, it is a midsegment.

This is just one of several converse theorems for the triangle midsegment theorem. We also prove another converse theorem - that if a line connecting two sides of a triangle is parallel to the third side and intersects one side’s midpoint, it is a midsegment.

Geometry shapes: a triangle with midsegment

Problem

In triangle ΔABC,  DE is parallel to BC, and its length is equal to half the length of BC. Show that DE is a midsegment. In other words: Show that BD=DA and CE=EA.

Strategy

Since this is a converse theorem, it is often a good strategy to solve the geometry problem by looking at how we proved the original theorem and do things in the opposite order.

We proved the original theorem here - Triangle Midsegment Theorem , and did so by constructing another triangle by extending DE to point F so that DE=EF. Let's do that here, as well:

Triangle with Midsegment extended (Geometry shapes)

Now, since DE=EF, and DE is half of BC, DF should be equal to BC. We now have a quadrilateral, DFCB, in which there is a pair of opposite sides (DF and BC) which are both parallel (given) and equal in length (we constructed DF that way), and as we have shown such a quadrilateral is a parallelogram.

If DFCB is a parallelogram, BD=CF, and BD||CF, as opposite sides of a parallelogram.

Continuing to work backward from the way we proved the original theorem, we will now show the triangles ΔADE and ΔCFE are congruent, and from this show that CE=EA, and also that CF=AD, and as a result BD=AD.

Proof

(1) DE=EF //Construction
(2) DE= ½BC       //Given
(3) EF = ½BC       //(1), transitive property of equality
(4) DE+EF= ½BC+ ½BC //Add (2) and (3)
(5) DF = BC //Simplify (4)
(6) DE||BC //Given
(7) DFCB is a parallelogram //(5), (6), quadrilateral with two sides parallel and equal
(8) BD=CF //(7), Opposite sides of parallelogram
(9) BD||CF //(7), Opposite sides of parallelogram
(10) ∠DAE ≅ ∠FCE   // Alternate Interior Angles Theorem
(11) ∠AED ≅ ∠CEF    // Vertical angles
(12) DE=EF //Construction
(13) ΔADE ≅ ΔCFE // Angle-Angle-Side
(14) CE=EA // corresponding sides in congruent triangles, (CPCTC)
(15) CF=DA // corresponding sides in congruent triangles, (CPCTC)
(16) CF=BD // (8)
(17) DA=BD // (15), (16), transitive property of equality

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About the Author

Ido Sarig is a high-tech executive with a BSc degree in Computer Engineering. His goal is to help you develop a better way to approach and solve geometry problems. You can contact him at [email protected]

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Welcome to Geometry Help! I'm Ido Sarig, a high-tech executive with a BSc degree in Computer Engineering and an MBA degree in Management of Technology. I'm here to tell you that geometry doesn't have to be so hard! My goal with this website is to help you develop a better way to approach and solve geometry problems, even if spatial awareness is not your strongest quality. Read More…

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